Please help.............

Question

A 25.0-mL sample of 0.100 M propanoic acid (HC3H5O2, Ka = 1.3 multiplied by 10-5) is titrated with 0.100 M NaOH solution. Calculate the pH after the addition of the following amounts.

(a) 0.0 mL

(b) 4.0 mL

(c) 8.0 mL

(d) 12.5 mL

(e) 20.0 mL

(f) 24.0 mL

(g) 24.5 mL

(h) 24.9 mL

(i) 25.0 mL

(j) 25.1 mL

(k) 26.0 mL

(l) 28.0 mL

(m) 30.0 mL

Answer 1

a) [H+] = (Ka x Ma)^1/2

= (1.3x10^-5 x 0.1M)^1/2 = 1.14 x 10^-3

pH = 2.9

b) HC3H5O2 + NaOH → C3H5O2Na + H2O

2.5mmol...0.4mmol

0.4mmol...0.4mmol

----------------------------

2.1mmol

This is the buffer solution :

[H+] = Ka x mol of acid / mol of salt x val

= 1.3x10^-5 x 2.1mmol / 0.4 mmol x 1

= 6.825 x 10^-5

pH = 4.166

For the next number, i didn't write my steps. I hope you can do it as same as my steps in my answer above.

c) pH = 4.56

d) pH = 4.89

e) pH = 5.45

f) pH = 6.27

g) pH = 6.58

h) pH = 7.28

i) Use formula of Hydrolisis :

[OH-] = (Kw x [salt] x val /Ka)

From the formula, pH is found =  8.8

j) pH = 10.3

k) pH = 11.3

l) pH = 11.75

m) pH = 11.96